Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy sequences

  1. Jun 16, 2005 #1
    hello all

    I found this rather interesting
    suppose that a sequence [tex]{x_{n}}[/tex] satisfies

    [tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex] [tex] \forall n\epsilon N[/tex]

    how couldnt the sequence [tex]{x_{n}}[/tex] not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

  2. jcsd
  3. Jun 16, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    An obvious one: just add those differences. Let xn be [tex]\sum_{i=1}^n \frac{1}{n}[/tex]. That series does not converge and so the sequence of partial sums is not Cauchy.
  4. Jun 17, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

    It looks like a Cauchy sequence to me based on the definition:

    The sequence: [itex]\{x_n}\}[/itex] is a Cauchy sequence if given [itex]\epsilon[/itex] there exists N such that for all m,n[itex]\leq[/itex]N we have:


    For the sequence:

    [tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex]

    I take [itex]\epsilon=\frac{1}{N+1}[/itex];

    Thus for all n>N:

  5. Jun 17, 2005 #4
    That's true; but look closely - is that really the same thing as

  6. Jun 17, 2005 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
    You do remind me that just [tex]|x_{n+1}-x_n|< \epsilon[/tex] is not enough. You have to show that [tex]|x_n-x_m|< \epsilon[/tex] which is basically showing that the tail of the sum goes to 0.
  7. Jun 17, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    I suppose not.


    does not imply


    Still a little unclear. I'll work on it. Thanks guys.

    Seriously, why do I even bother with differential equations anyway . . .
  8. Jun 17, 2005 #7
    Simply put: for given N, your

    [tex] | x_m - x_n | [/tex],

    [tex] \mbox{ as } m \rightarrow \infty [/tex],

    is not guaranteed to be any better than the (divergent) harmonic series:

    [tex] | x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty [/tex]
  9. Jun 17, 2005 #8


    User Avatar
    Science Advisor
    Homework Helper

    Alright, I'm interested in getting my math right.
    Following Hall's example (a slight change):


    [tex]x_n=\sum_{i=1}^{n+1} \frac{1}{i}[/tex]


    [tex]|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}[/tex]



    However, the partial sums of this harmonic series [itex] \{x_n\}[/itex] do not converge and thus cannot be a Cauchy sequence.

    Now, I got a pot of spagetti to make . . .
  10. Jun 17, 2005 #9
    hello all

    nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i cant think of any other examples, do you know of any others?

  11. Jul 27, 2011 #10
    hi guys i need bounded but not cauchy function can you help me asap?
  12. Jul 28, 2011 #11
    Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Cauchy sequences
  1. Cauchy Sequence (Replies: 1)

  2. Cauchy sequence (Replies: 10)

  3. Cauchy Sequences (Replies: 4)

  4. Cauchy sequence (Replies: 6)

  5. Cauchy Sequences (Replies: 4)